Question

Find a formula for the polynomial of least degree that is graphed below Its x-intercepts are 0 and -3, and the graph goes through the point (-1,4).

Answer 1

The polynomial of least degree with the x-intercepts 0, -3 and passing through (-1,4) is f(x) = -2x^2 - 6x.

Step-by-step explanation:

To find the polynomial of least degree with given x-intercepts and passing through a specific point, we start by constructing a polynomial function using its intercepts. Since the x-intercepts are 0 and -3, the polynomial must have factors of x (because the graph crosses the y-axis at 0) and (x + 3) (because it crosses the x-axis at -3). A polynomial of least degree with these intercepts would be a quadratic polynomial of the form:

f(x) = ax(x + 3)

Next, to determine the correct value of a, we use the given point (-1,4). Substituting x with -1 gives us:

f(-1) = a(-1)(-1 + 3) = 4

This simplifies to:

-2a = 4

So, a = -2. Therefore, the required polynomial is:

f(x) = -2x(x + 3) = -2x2 - 6x

This is the polynomial equation of least degree that satisfies the given conditions.

Answer 2

Answer: y=-2x²-6x

Step-by-step explanation:

Its x-intercepts are 0 and -3, and the graph goes through the point (-1,4)

Hence we have three coordinates of the equation y=ax²+bx+c (1):

(0,0) (-3,0) (-1,4)

Let's substitute these coordinates into formula (1):

`0=a(0)^2+b(0)+c\\0=a(-3)^2+b(-3)+c\\4=a(-1)^2+b(-1)+c\\\\0=0+0+c\\0=9a-3b+c\\4=a-b+c\\\\0=c\\0=9a-3b+0\\0=a-b+0\\\\c=0\\9a-3b=0\ \ (1)\\a-b=4\ \ \ \ \ (2)\\\\c=0\\\\`

Divide both parts of the equation (1) by 3 and multiply both parts of the equation (2) by -1: :

`3a-b=0\\-a+b=-4\\\\`

Sum these equations:

`2a=-4`

Divide both parts of the equation by 2:

`a=-2`

We substitute the value of a=-2 into equation (2):

`-2-b=4\\`

Divide both parts of the equation by -1:

`2+b=-4\\\\2+b-2=-4-2\\\\b=-6`

`Thus, \\\\y=-2x^2-6x+0\\\\y=-2x^2-6x`